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物理代写|量子力学代写Quantum mechanics代考|Probability Amplitudes for Composite Systems

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物理代写|量子力学代写Quantum mechanics代考|Probability Amplitudes for Composite Systems

We relied on the orthogonality of the two-qubit computational basis states for evaluating amplitudes such as $\langle 00 \mid 10\rangle$ or $\langle 00 \mid 00\rangle$ in the above matrix representation. It turns out that there is another way to evaluate these amplitudes that relies only on the orthogonality of the single-qubit computational basis states.

Suppose that we have four single-qubit states $\left|\phi_0\right\rangle,\left|\phi_1\right\rangle,\left|\psi_0\right\rangle,\left|\psi_1\right\rangle$, and we make the following two-qubit states from them:
$$\left|\phi_0\right\rangle \otimes\left|\psi_0\right\rangle, \quad\left|\phi_1\right\rangle \otimes\left|\psi_1\right\rangle$$
We may represent these states equally well as follows:
$$\left|\phi_0, \psi_0\right\rangle, \quad\left|\phi_1, \psi_1\right\rangle$$
because the Dirac notation is versatile (virtually anything can go inside a ket as long as its meaning is not ambiguous). The bra $\left\langle\phi_1, \psi_1\right|$ is dual to the ket $\left|\phi_1, \psi_1\right\rangle$, and we can use it to calculate the following amplitude:
$$\left\langle\phi_1, \psi_1 \mid \phi_0, \psi_0\right\rangle$$

物理代写|量子力学代写Quantum mechanics代考|Controlled Gates

An important two-qubit unitary evolution is the controlled-NOT (CNOT) gate. We consider its classical version first. The classical gate acts on two cbits. It does nothing if the first bit is equal to zero, and flips the second bit if the first bit is equal to one:
$$00 \rightarrow 00, \quad 01 \rightarrow 01, \quad 10 \rightarrow 11, \quad 11 \rightarrow 10$$
We turn this gate into a quantum gate ${ }^5$ by demanding that it act in the same way on the two-qubit computational basis states:
$$|00\rangle \rightarrow|00\rangle, \quad|01\rangle \rightarrow|01\rangle, \quad|10\rangle \rightarrow|11\rangle, \quad|11\rangle \rightarrow|10\rangle .$$
By linearity, this behavior carries over to superposition states as well:
$$\alpha|00\rangle+\beta|01\rangle+\gamma|10\rangle+\delta|11\rangle \quad \stackrel{\text { CNOT }}{\longrightarrow} \alpha|00\rangle+\beta|01\rangle+\gamma|11\rangle+\delta|10\rangle .$$
A useful operator representation of the CNOT gate is
$$\mathrm{CNOT} \equiv|0\rangle\langle 0|\otimes I+| 1\rangle\langle 1| \otimes X .$$
The above representation truly captures the coherent quantum nature of the CNOT gate. In the classical CNOT gate, we can say that it is a conditional gate, in the sense that the gate applies to the second bit conditioned on the value of the first bit. In the quantum CNOT gate, the second operation is controlled on the basis state of the first qubit (hence the choice of the name “controlled-NOT”). That is, the gate acts on superpositions of quantum states and maintains these superpositions, shuffling the probability amplitudes around while it does so. The one case in which the gate has no effect is when the first qubit is prepared in the state $|0\rangle$ and the state of the second qubit is arbitrary.

物理代写|量子力学代写Quantum mechanics代考|Probability Amplitudes for Composite Systems

$$\left|\phi_0\right\rangle \otimes\left|\psi_0\right\rangle, \quad\left|\phi_1\right\rangle \otimes\left|\psi_1\right\rangle$$

$$\left|\phi_0, \psi_0\right\rangle, \quad\left|\phi_1, \psi_1\right\rangle$$

$$\left\langle\phi_1, \psi_1 \mid \phi_0, \psi_0\right\rangle$$

物理代写|量子力学代写Quantum mechanics代考|Controlled Gates

$$00 \rightarrow 00, \quad 01 \rightarrow 01, \quad 10 \rightarrow 11, \quad 11 \rightarrow 10$$

$$|00\rangle \rightarrow|00\rangle, \quad|01\rangle \rightarrow|01\rangle, \quad|10\rangle \rightarrow|11\rangle, \quad|11\rangle \rightarrow|10\rangle .$$

$$\alpha|00\rangle+\beta|01\rangle+\gamma|10\rangle+\delta|11\rangle \quad \stackrel{\text { CNOT }}{\longrightarrow} \alpha|00\rangle+\beta|01\rangle+\gamma|11\rangle+\delta|10\rangle .$$
CNOT门的一个有用的算子表示是
$$\mathrm{CNOT} \equiv|0\rangle\langle 0|\otimes I+| 1\rangle\langle 1| \otimes X .$$

MATLAB代写

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